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A-LevelPhysicsCapacitance9702/41 · May/June 2021 · Q79 Marks

State what is meant by the capacitance of a parallel plate capacitor.

  1. (a)State what is meant by the capacitance of a parallel plate capacitor.[2]
  2. (b)A capacitor of capacitance C is connected into the circuit shown in [Figure 7.1]. When the two-way switch is in position A, the capacitor is charged so that the potential difference across it is V. The switch moves to position B and the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f.
  3. (b)(i)Show that the average current I in the ammeter is given by the expression I = fCV.[2]
  4. (b)(ii)For a potential difference V of 150 V and a frequency f of 60 Hz, the average current in the ammeter is 4.8 µA. Calculate the capacitance, in pF, of the capacitor. capacitance = ........................................................... pF [2][2]
  5. (c)A second capacitor, having the same capacitance as the capacitor in (b), is connected into the circuit of [Figure 7.1]. The two capacitors are connected in series. State and explain the new reading on the ammeter. new reading = ........................................................... µA [3][3]

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Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

  1. [1] (a) charge / potential
  2. [1] charge is on one plate, potential is p.d. between the plates
  3. [1] (b)(i) I = Q/t
  4. [1] charge = CV and time = 1 / f leading to I = fCV
  5. [1] (b)(ii) 4.8 x 10^-6 = 150 × 60 × C
  6. [1] C = 530 pF
  7. [1] (c) (total) capacitance is halved
  8. [1] charge (for each cycle/discharge) is halved
  9. [1] since f and V are constant, current is proportional to capacitance
  10. [1] current = 2.4 μA

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About this question

Topic
Capacitance, A-Level Physics (syllabus 9702). Structured question, 9 marks.
Source
A-Level Physics, May/June 2021 Paper 4 Variant 1, question 7.

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